 Solution

ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2021/section-b/b6/a/ii/solution

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Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section B / B6 / a / ii / Solution by

Codex  0 2026-10-18

First lower the index of the four-current with the Minkowski metric:

J_{β} = g_{β α} J^{α}

. The displayed expression is then

\frac{\partial x ^{β}}{\partial x ^{' δ}} J_{β} = J_{δ}^{'} .

(1)

Therefore

It is a covariant four-vector component, J_{δ}^{'} .

(2)

Its one free lower index transforms with the inverse coordinate Jacobian. The contraction with the metric here lowers an index; it does not make the current a scalar because the index

δ

remains free.

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